﻿#include "header.h"

/*

给定一个单词数组和一个长度 maxWidth，重新排版单词，使其成为每行恰好有 maxWidth 个字符，且左右两端对齐的文本。
你应该使用“贪心算法”来放置给定的单词；也就是说，尽可能多地往每行中放置单词。必要时可用空格 ' ' 填充，使得每行恰好有 maxWidth 个字符。
要求尽可能均匀分配单词间的空格数量。如果某一行单词间的空格不能均匀分配，则左侧放置的空格数要多于右侧的空格数。
文本的最后一行应为左对齐，且单词之间不插入额外的空格。

说明:
单词是指由非空格字符组成的字符序列。
每个单词的长度大于 0，小于等于 maxWidth。
输入单词数组 words 至少包含一个单词。

示例 1:
输入:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
输出:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]
		 
示例 2:
输入:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
输出:
[
	"What   must   be",
	"acknowledgment  ",
	"shall be        "
]
解释: 注意最后一行的格式应为 "shall be    " 而不是 "shall     be",
因为最后一行应为左对齐，而不是左右两端对齐。第二行同样为左对齐，这是因为这行只包含一个单词。

示例 3:
输入:
words = ["Science","is","what","we","understand","well","enough","to","explain",
		 "to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
输出:
[
	"Science  is  what we",
    "understand      well",
	"enough to explain to",
	"a  computer.  Art is",
	"everything  else  we",
	"do                  "
]
						
*/

class Solution {
public:
	vector<string> fullJustify(vector<string> &words, int maxWidth)
	{
		vector<string> res;
		for (int i = 0, k = 0, count = 0; i < words.size(); i += k, k = 0, count = 0)
		{
			//单词长度除了本身外，还应该再加一个空格。
			while (i + k < words.size() && count + words[i + k].size() <= maxWidth - k)
			{
				count += words[i + k].size();
				++k;
			}
			string temp = words[i];
			for (int j = 0; j < k - 1; ++j)
			{
				if (i + k >= words.size())//最后一行
					temp += " ";
				else
					temp += string((maxWidth - count) / (k - 1) + (j < (maxWidth - count) % (k - 1)), ' ');
				temp += words[i + j + 1];
			}
			temp += string(maxWidth - temp.size(), ' ');
			res.push_back(temp);
		}
		return res;
	}
};
//https://leetcode.com/problems/text-justification/discuss/24873/Share-my-concise-c%2B%2B-solution-less-than-20-lines